Animated Solution for Mathematics - Definite Integration: The area of the region described by $A=\{(x,y):x^2+y^2\le 1\text{ and }{y}^2\le 1-x\}$ is:

Imagine you are standing on the Cartesian plane, looking at two distinct mathematical entities. The first is a perfect, symmetric circle defined by $x^2+y^2\le 1$. This is our familiar unit circle, centered at the origin, a shape of pure elegance.

The second is a more dynamic, sweeping curve: the parabola $y^2\le 1-x$. This parabola has its vertex at $(1,0)$ and opens aggressively to the left.

Our goal is to find the area of the region $A$ where these two shapes overlap. This is not just a calculation; it is a journey of understanding how different functions compete to define a space.

Solving the Algebra

To understand the bounded region, we must first find where these two curves meet. We set their boundary equations equal: $x^2+y^2=1$ and $y^2=1-x$.

By substituting the parabola's definition of $y^2$ directly into the circle's equation, we get:

The ones on both sides cancel out, leaving us with the beautiful, simple equation $x^2-x=0$. Factoring this, we find $x(x-1)=0$, which gives us the intersection points at $x=0$ and $x=1$.

When $x=0$, $y^2=1$, so $y=\pm 1$. When $x=1$, $y^2=0$, so $y=0$. These points—$(0,1)$, $(0,-1)$, and $(1,0)$—are the anchors of our region.

Why We Split the Area

Now, look at the graph. The $y$-axis ($x=0$) acts as a natural divider. For $x\le 0$, the circle is the boundary that keeps our region contained.

But as soon as we cross into the positive $x$ territory, the parabola takes over as the tighter constraint. Because the function defining the boundary changes, we must split our total area into two parts: $A_1$ for the left side and $A_2$ for the right side.

The total area is simply $A=A_1+A_2$.

For $A_1$, the region where $x\le 0$, we don't need complex calculus. It is a perfect semi-circle with radius $r=1$.

The area of a full circle is $\pi r^2$, so our semi-circle is simply:

For $A_2$, the region where $x\ge 0$, we turn to integration. The region is bounded by the parabola $y^2=1-x$. Because of the symmetry across the $x$-axis, we can integrate the upper branch $y=\sqrt{1-x}$ from $x=0$ to $x=1$ and multiply by 2.

Our integral is:

Using the power rule, the integral of $(1-x{)}^{1/2}$ is $-\frac{2}{3}(1-x{)}^{3/2}$. Applying the limits from 0 to 1, we get:

At $x=1$, the term is 0. At $x=0$, the term is $-\frac{2}{3}$. Subtracting these gives $2(0-(-2/3))=\frac{4}{3}$.

We have arrived at the finish line. We have the area of the semi-circle, $\frac{\pi }{2}$, and the area under the parabola, $\frac{4}{3}$.

Adding these two components together, the total area of the region $A$ is:

This matches option (2). You have successfully navigated the intersection of geometry and calculus, proving that even complex problems can be broken down into elegant, manageable steps.